You can put a vector in 3 dimensional space and draw a vector (directed line) to that point P(x & y & z) which turns into the vector v in .
v is represented as
notation says this is the length of vector v. That symbol is called “norm”.
Ex: v and w are two vectors, so to sum them, we create a parallelogram and calculate the width(?) of the resulting parallelogram.
To subtract, we build a triangle. iff v and w have the same direction and the same length.
The distance between and is This is the same as:
|| (v_1 - w_1) ~~ (v_2-w_2) ~~ (v_3-w_3) || \end{aligned} $$ $=\sqrt{(v_1-w_1)^2+(v_2-w_2)^2+(v_3-w_3)^2}$ Example: Find: the distance between $\underbar{P}_1(2,-1,3)$ and $\underbar{P}_2(1,1,4)$. Always subtract the second one from the first one. $\overrightarrow{\underbar{P}_1 \underbar{P}_2} = \underbar{P}_2 - \underbar{P}_1$ $$ \begin{bmatrix} 1-2\\ 1--1\\ 4-3 \end{bmatrix} $$ Which becomes $$ \begin{bmatrix} -1\\ 2\\ 1 \end{bmatrix} $$ ## Scalar multiple If a is a real number and $v \ne 0$ is a vector, then: 1. $||av|| = |a| ||v||$ (absolute value of a times the length of v) 2. If $av \ne 0$, the direction of $av$ is: a. the same as v if a > 0 b. opposite to v if a < 0 ## Unit vector :drill: A vector is a unit vector if it has length [1], which means the norm is equal to [1]. If $v\ne0$ then $\frac{\vec{v}}{||v||}$ is a unit vector. Since: $|| \frac{v}{||v||} ||$ is the same as: $$ \begin{aligned} ||\frac{1}{||v||} \cdot ||v|| || \end{aligned} $$ and we know that we can pull that out from the scalar multiple line. Then it turns into $\frac{||v||}{||v||}$ which is equal to 1. Given a non-zero vector v, normalize = take the unit vector $\frac{\vec{v}}{||v||}$. Normalize $v=\begin{bmatrix}0\\ 4\\ -3\end{bmatrix}$ Sol: $$ \begin{aligned} ||v|| &= \sqrt{0^2 ~~ 4^2 ~~ -3^2}\\ &= 5 \end{aligned} $$ so $$ \begin{aligned} \frac{v}{||v||} = \frac{1}{5}\begin{bmatrix}0\\ 4\\ -3\end{bmatrix}\\ = \begin{bmatrix}0\\ \frac{4}{5}\\ -\frac{3}{5} \end{bmatrix}\\ \end{aligned} $$ -- ## Parallel :drill: Two nonzero vectors re called parallel if they have the same or opposite [direction]. As a consequence, two non-zero vectors are parallel (symbol is: ||) iff one is a scalar multiple of another. ## Direction Vector :drill: A vector that is parallel (||) to any two distinct vectors on the line. I think this is really saying: Given two distinct vectors, there is a third vector which goes through a different point on each line. The direction vector is parallel to that vector. ## Lines There is one line that passes through a particular point $\underbar{P}_0(x_0,y_0,z_0)$ and has a direction vector $d = \begin{bmatrix}a\\ b\\ c\end{bmatrix}$. This means that $\overrightarrow{P_0P}||d$ (vector from $P_0$ to P is parallel to d) iff $\overrightarrow{P_0P} = td$ for $t \in \mathbb{R}$. $$ \begin{aligned} \begin{bmatrix}x-x_0\\ y-y_0\\ z-z_0\end{bmatrix} = t\begin{bmatrix}a\\ b\\ c\end{bmatrix} \\ \begin{cases} x = x_0+ta\\ y=y_0+tb\\ z=z_0+tc\\ \end{cases} \end{aligned} $$ -- $$ \begin{aligned} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} x_0\\ y_0\\ z_0 \end{bmatrix} + t \begin{bmatrix} a\\ b\\ c \end{bmatrix} \end{aligned} $$ is the line through $\underbar{P}_0$ in the direction d that is: $$ \begin{aligned} \begin{bmatrix}x-x_0\\ y-y_0\\ z-z_0\end{bmatrix} = t\begin{bmatrix}a\\ b\\ c\end{bmatrix} \end{aligned} $$ for $t \in \mathbb{R}$ ## Parametric equations of a line The line through $\underbar{P}_0(x_0,y_0,z_0)$ with direction $d = \begin{bmatrix}a\\ b\\ c\end{bmatrix} \ne 0$ is given by: $$ \begin{cases} x=x_0+ta\\ y=y_0+tb\\ z=z_0+tc\\ \end{cases} $$ EX: Find the equations of the line which pass through $\underbar{P}_0(2,0,1)$ and $\underbar{P}_1(4,-1,1)$ Sol: $d = \overrightarrow{P_0P_1}$ $$ \begin{bmatrix}x_1-x_0\\ y_1-y_0\\ z_1-z_0\end{bmatrix} = \begin{bmatrix}2\\ -1\\ 0\end{bmatrix} $$ (which is d) The line is: x = 2+2t y = 0+t z = 1 Works out to have: $\underbar{P}_0$ coords - d$ in equation form. Ex: Determine whether the following lines intersect and, if so, find the point of intersect. x=1-3t y=2+5t z=1+t x=-1+s y=3-4s z=1-s Sol: Let x,y,z be the point of intersection, ie suppose P(x,y,z) lies on both lines. The lines intersect iff 1-3t = -1+s 2+5t = 3-4s 1+t = 1-s $$ \begin{cases} s+3t=2\\ -4s-5t=-1\\ -s-t=0\\ \end{cases} $$ bottom one says s= -t $$ \begin{cases} t=1\\ s=-1\\ s=-t \end{cases} $$ So the point of intersection is: t: 1-3 = -2 2+5 = 7 1+1 = 2 s: -1+-1 = -2 3+4 = 7 1--1 = 2